Squeezes, Part III. Problems with "L"

[inquiry]

This thread will continue the topic of identifying squeezes at the table based upon pattern recognition. It is assumed that you have read or at least know on your own the material covered in the first two threads dealing with squeezes and pattern recognition....

 

First thread was "Introduction to squeeze plays, and the second thread was Squeezes, Part II

 

In the first thread we learned how to identify the four cardinal requirements of a squeeze, namely BLUE. In addition, we spent a fair amount of time on the necessary entry condition in the first thread for and how to compensate for flawed entry requirement.

 

In the second thread, we learned how to deal with problems with the "BOTH" requirement for a simple squeeze. This included a large number of squeeze that share this single defect, and problems (just like in chapter one) that crop up with changing entry conditions.

 

In this thread we will cover squeezes where there is a problem with "L" or the number of losers. We will actually examine squeezes where "L" might be viewed as zero, but that a squeeze is needed to get all your tricks. We will see some blocked entry conditions where "L" might be viewed as one, but you need a squeeze just to lose only one trick, but mostly we will look at squeezes where "L" is two or greater.

 

There are many different ways you can a have a defect in the number of losers. The most common is to have two or more losers when you are trying to win just one extra trick. It turns out that defects in the “loser” requirement can be a blessing. But we will start with the cardinal rule of hands where "L" is greater than one and you are thinking squeeze (to paraphrase Cyde B Love who said this or something like this). "When the loser count is two and you are thinking of a squeeze, as soon as it is safe to do so, lose the trick you need to lose."

 

This advice from Professor Love is perhaps the easiest remedy and is the easiest way to “correct the count” for the squeeze. There is not a lot of strategy in most of the correct the count positions. You lose the trick(s) necessary to get the count down to where the squeeze can be executed. Sometimes, complications will not let you correct the count at first (they will knock out your entry with a shift, they will get a ruff, they will immediatly cash a second trick, etc.

 

We have already seen some "correct the count" hands in the other threads. For instance, at least Rebound's hand in the Basic Squeeze Quiz thread along with quiz number five deal with correcting the count.

 

Next time. Squeezes where Losers = 0, and squeezes where Loser = 1 but you have to lose a trick and squeeze to avoid losing two tricks (blocked endings).

 

ben

[inquiry]

Unusual Problem with "L", zero losers and one loser where you are willing to lose a trick

 

The squeezes we have examined so far require the loser count to be one. As noted in the introduction to this threat, the next defect in the basic squeeze position we shall examine is when the loser count is defective. We will shortly turn to plays to help you compensate for endings in which the loser count is too high (L=2+), but bizarrely enough you will sometimes need a squeeze plays to help you compensate for when the defect is that you loser count is not too high (say 2 losers instead of 1), but rather it is too low! That's right, it is sometimes necessary to execute a squeeze play when your loser count is zero!!!

 

How can this be, you might ask? With zero losers your first impulse might be to assume you can simply claim all the remaining tricks. But after the last two threads, you now realize that blocked entry conditions may exist that make it impossible to simply cash all your winners. This is when a squeeze play can come to your rescue. We will examine just such a blocked ending, and then, for completeness, show a similar ending where the loser count is one but due to the blocked entry condition you will be at risk of losing two tricks. In this second case, you will not be trying to squeeze to gain the extra trick, but rather to just get what is rightfully yours.

 

[hv=n=shkdcat&w=shadcqj&e=shdimmaterialc&s=sah2dck]399|300|Proof of principle: Blocking ending squeeze with L = 0

 

You have two ♣ and one ♠ winner but no way to cash your ♣king and then get to dummy for the ♣A. But this is like a clash squeeze on just one person. He has little choice but to give you your third trick when cash the ♠Ace. OF course during play, you would try to avoid such blockage if you can. [/hv]

 

That squeeze is neat, but hardly anything to get excited about. But how about an ending like this one?

[hv=n=shkdcat&w=shadcqj&e=shdimmaterialc&s=sah2dck]399|300|Proof of principle: Blocking ending squeeze with L = 0

 

You have two ♣ and one ♠ winner but no way to cash your ♣king and then get to dummy for the ♣A. But this is like a clash squeeze on just one person. He has little choice but to give you your third trick when cash the ♠Ace. OF course during play, you would try to avoid such blockage if you can. [/hv]

 

In the entry squeeze above, when you play a spade to dummy, what is EAST to play? IF it is a heart, one ♥ hook is enough for 3♥ tricks. If it is a ♣, your club is good in dummy and two ♥'s will be enough. But what if he discards a ♦? Here is what an entry squeeze does for you. The ♦ discard doesn't really set up an immediate trick for you, you are still going to win the two diamonds you already had. It looks like this is no squeeze at all because of that, but watch what happens, if by magic. You now lead the diamond Queen to the King!! Take the heart hook, and then lead the ♦2 to the ♦3 and repeat the heart hook.

 

This is one of many "non-material squeezes" covered by Ottik and Kelsey in "Adventures and in card play", my absolutely favorite book. You have little chance of getting these ending right if you have never studied how to deal with block entry condition squeezes. This entry squeeze will turn up again later when you have a sure losers but a blocked position like this.

 

What have we learned

• entry squeeze may give you an entry when you are an entry short to unravel your winners
  
• Some squeeze work with an apparent loser count of zero but that block condition prevents you from winning all your tricks
  
• Adventrures in card play in inquiries favorite bridge book, nah, any kind of book
  

Next time Bigger Loser counts.

[inquiry]

Loser count = 2, you need all the rest of the tricks: Part A. The triple squeeze

 

We have learned in the first two threads that when the loser count = 1, a squeeze can very often force another winner for you. But we were careful to keep loser = 1. When loser = 2, it is possible, sometimes, for a squeeze to either gain two tricks immediately, or set up a winner that when cashed excutes another squeeze. These two trick gaining squeezes are often referred to as "progressive", and we shall see a variety of them that work either against one opponent, or against both.

 

Let's start with the easy progressive squeeze to find and to execute, the so called triple squeeze (there is a non-progressive version as well, which gains only one trick).

 

We have seen a large number of hands where one opponent was squeeeze in three suits (such as compound squeeeze, guard squeeze, clash squeeze), and we found that this squeeze worked on the next to last free winner, and you still had a squeeze card left to play. In those endings, L=1, so there was room for one more squeeze card. In a two loser squeeze, you will not have that extra squeeze card, but the last squeeze card can setup an extra winner. Let's look at a very simple ending...

[hv=n=shajdajcxx&w=shkqdkqckq&e=sht9dt9ct9&s=sahxxdxcaj]399|300|Winners, four aces, so L = 2.

 

West alone protects against all three jacks. Heart ACE is the primary entry.

 

When you cash the spade ACE, West is squeezed in three suits. If he throws a club, you win ♣A then ♣JACK to squeeze him in the red suits.

 

If he throws a ♥, you win the ♥A then jack to squeeze him in the minors. And if he throws a ♦You win diamond ace and a diamond to squeeze him in the rounded suits. [/hv]

 

This squeeze is progressive, the spade ACE forces WEST to discard a guard which sets up a second squeeze. For the second squeeze to work, BLUE has to be satisfied. This suggest that if WEST has an option which winner to set up for you he can do so that either A.) forces you to use your primary entry so the second squeeze will not work, or B.) sets up the winner in a suit that you no longer have an UPPER threat.

 

Let's make a minor change to the last hand, to see one time when a triple squeeze is not progressive.

[hv=n=shajdajcxx&w=shkqdkqckq&e=sht9dt9ct9&s=sahxxdxcaj]399|300|Winners, four aces, so L = 2.

 

West alone protects against all three jacks. Heart ACE is the primary entry.

 

When you cash the spade ACE, West is squeezed in three suits. If he throws a club, you win ♣A then ♣JACK to squeeze him in the red suits.

 

If he throws a ♥, you win the ♥A then jack to squeeze him in the minors. And if he throws a ♦You win diamond ace and a diamond to squeeze him in the rounded suits. [/hv]

 

You should see why the squeze repeated in the first instance (two threats in the upper hand) and it didn't repeat in the second case.

 

What we have learned.

• When loser is flawed such that :L = 2, the squeeze may still gain two tricks if one opponent holds guards in all thre suits
  
• This repeating triple squeeze requires necesssary entries, and two threats in the upper hand
  
• A good defensive measure to break up the three suit squeeze is to establish the threat in the upper hand
  

Next time, ways to overcome the lack of TWO threats in the upper hand or entry problems.

[inquiry]

Loser =2, only one threat in the upper hand

 

Last time we examined a squeeze that would gain TWO tricks if 1) only one hand was guarding against threats in three suits, 2) if the entries were right, and 3) if two of the three threat suits lay in the upper hand.

 

This last requirement is easy to see,once you figure out who you are squeezing. Your victim can escape the squeeze for the second trick by establishing the threat in the upper hand if the upper hand had one threat. That is simple. The solution that often repairs this problem is also simple if you think about it. What would that be? If the person you squeeze has to give up two tricks immediately if he establishes the threat in the upper hand. When will this gift occur? When the threat in the upper hand is what is often referred to as an "extended threat".. here is an example.

 

[hv=n=shxxdaj3cx&w=shkqdkqckq&e=sht9dt9ct9&s=sahajdxcaj]399|300|This ending should look familiar...this is almost exactly the losing position at the end of the post above, but notice, we move one small club from north's hand to diamonds. What a difference that 3 of diamonds makes. When you cash your spade ace now, West is deprived of his killing defense before of establishing diamonds. You can see why now. If he discards a diamond, you win not one extra diamond trick (the JACK), but two (the diamond 3 will be established too). [/hv]

 

Extended threats will be useful in a lot of multiple loser squeezes. Be sure to learn to look for them, and not to squander small cards in one of your potential threat suits if you can help it.

 

What have we learned?

• What an extended threat
  
• In a triple squeeze with only one threat in the upper hand, if that threat is an extended threat, the squeeze will be progressive